What would be the length of a day(that is, the time required forone
rotation of the earth on its axis ) if the rate of rotation ofthe
earth were such that g=0 at the equator?

T = 5,06 10³ s

Explanation:

For this exercise, let's use Newton's second law, where force is the force of gravitational attraction

            F = m a

The acceleration is centripetal

          a = v² / r

Let's replace

         G m / ² = m v² /

         G / = v²

The velocity module is constant, so we can use the equation of uniform motion

        v = d / t

Where the distance  is the length of the circle and in this case the time is called the period

         d = 2π

We replace

     (2π / T)² = G /

       T² = 4π² ³ / G

        T = √ 4π² ³ / G

Let's calculate

        T = √ (4π² (6.37 10⁶)³ / 6.67 10⁻¹¹ 5.98 10²⁴)

        T = √ (255.83 10⁵) = √ (25.583 10⁶)

        T = 5,05796 10³ s

for this one it would be d bump/forearm pass